Question

If $a$ and $b$ are real numbers between $0$ and $1$ such that the points $(a,1),(1,b)$ and $(0,0)$ from an equilateral triangle, then $a=b=2-\sqrt{3}$.

Answer 1

${(a-1)}^2+{(1-b)}^2=a^2+1=b^2+1={\left(side\right)}^2$

The last two imply $a=b$ or $a=-b$

The first two imply $b^2-2b-2a+1=0$

If $a=-b$, then $b^2+1=0$ not possible

If $a=b$, then $b^2-4b+1=0$

$\therefore b=2+\sqrt{3},2-\sqrt{3}$

Since $b<1\therefore b=2-\sqrt{3}=a$